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The coordinates of a particle of mass 'm...

The coordinates of a particle of mass `'m'` as function of time are given by `x=x_(0)+a_(1) cos(omegat)` and `y=y_(0)+a_(2)sin(omega_(2)t)`. The torque on particle about origin at time `t=0` is :

A

`m(-x_(0)b-y_(0)a)omega_(1)^(2)hatk`

B

`-m(x_(0)bomega_(2)^(2)-y_(0)a)omega_(1)^(2)hatk`

C

Zero

D

`+my_(0)aomega_(1)^(2)hatk`

Text Solution

Verified by Experts

The correct Answer is:
D
`vecr=(x_(0)+a_(1)cos(omega_(1)t)hati+(y_(0)+a_(2)sin(omega_(2)t))hatj`
`vecv=-a_(1)omega_(1)sin(omega_(1)t)hati+a_(2)omega_(2)cos(omega_(2)t)hatj`
`veca=-a_(1)omega_(1)^(2)cos(omega_(1)t)hati-a_(2)omega_(2)^(2)sin(omega_(2)t)hatj`
at `t=0`, `vecr=(x_(0)+a_(1))hati+y_(0)hatj`
`vecF=mveca=-ma_(1)omega_(1)^(2)hati`
`vectau=vecrxxvecF=(ma_(1)omega_(1)^(2)y_(0))hatk`.
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